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Show existence of $\epsilon > 0, \delta > 0$ such that for all $A\in U_\epsilon(Id_n)$ exists unique $B \in U_\delta(Id_n)$ with $B^2=A$.

$U_\epsilon(Id_n):=\{X \in M_{n\times n}| \Vert X-Id_n \Vert < \epsilon; \Vert \cdot \Vert \text{ Operator norm}\}$

Implicit function method: Define $F:(A,B) \mapsto B^2-A$ and noticing that $(Id_n,Id_n)$ is a solution, and also $\partial_{B}F|_{(Id_n,Id_n)}=2Id_n \in GL(n)$. Impl.f.th. guarantees an existence of unique function $$g:U_\epsilon(Id_n)\to U_\delta(Id_n)$$ such that $g(A)^2-A=0$. Which is equivalent to the statement if we set $B:=g(A)$.

Now I would like to show same using inverse function theorem. Define $f(B)=B^2$, then its differential is $df_{Id_n}=2Id_n$ invertible. Therefore we get open neighborhoods $U,V$ of $Id_n$ and $f(Id_n)=Id_n$ respectively with $f|_{U}:U\to V$ diffeomorphism.
In particular, bijectivity implies $\forall A\in V: \exists! B \in U: f(B)=B^2=A$

Question: How to shrink $U,V$ to the form of open balls?

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Since $V$ is open, you can find a ball $B_1$ of radius $\delta$ around the identity that is contained in $V$. Now, look at the preimage $X=f^{-1}(B_1)\subseteq U$. It is open, since $f$ is continuous. Therefore you can find a ball $B_2$ of radius $\epsilon$ around the identity that is contained into $f^{-1}(B_1)$ therefore into $U$. These should do the trick.

PS: I've used here that the open balls form a basis of your topology around $Id_n$.

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